The area of the region bounded by two curves f(x) and g(x) is with poins of intersection at x = a and x = b is given by:
[tex]Area= \int\limits^b_a {\left(f(x)-g(x)\right)} \, dx [/tex]
Given the curves [tex]y=x+\ln x[/tex] and [tex]y=x^3-x[/tex], the area bounded by the curves is given by:
[tex]Area= \int\limits^{1.507397}_{0.447141} {(x+\ln x)-(x^3-x)} \, dx \\ \\ =\int\limits^{1.507397}_{0.447141} {(2x+\ln x-x^3)} \, dx \\ \\ =\left[x^2+x\ln x-x- \frac{1}{4}x^4 \right]^{1.507397}_{0.447141} \\ \\ =\left[(1.507397)^2+1.507397\ln1.507397-1.507397- \frac{1}{4} (1.507397)^4\right] \\ -\left[(0.447141)^2+0.447141\ln0.447141-0.447141- \frac{1}{4} (0.447141)^4\right] \\ \\ =0.092686-(-0.617095)=0.709781\approx\bold{0.710}[/tex]
The x-cordinate of the centroid of the area bounded by the two curves is given by:
[tex]C_x= \frac{\int\limits^{1.507397}_{0.447141} {x\left[(x+\ln x)-(x^3-x)\right]} \, dx}{Area} \\ \\ =\frac{\int\limits^{1.507397}_{0.447141} {\left(2x^2+x\ln x-x^4\right)} \, dx}{0.709781} \\ \\ = \frac{\left[ \frac{2}{3} x^3+ \frac{1}{2}x^2\ln x- \frac{1}{4} x^2- \frac{1}{5} x^5 \right]^{1.507397}_{0.447141}}{0.709781} \\ \\ = \frac{0.625068-(-0.0744211)}{0.709781} \\ \\ = \frac{0.6994891}{0.709781} \approx0.985[/tex]
The y-cordinate of the centroid of the area bounded by the tw.o curves is given by:
[tex]C_y= \frac{\int\limits^{1.507397}_{0.447141} {\left[(x+\ln x)^2-(x^3-x)^2\right]} \, dx}{2(Area)} \\ \\ =\frac{\int\limits^{1.507397}_{0.447141} {\left(2x\ln x+(\ln x)^2-x^6+2x^4\right)} \, dx}{2(0.709781)} \\ \\ = \frac{\left[x^2\ln x- \frac{1}{2} x^2+x(\ln x)^2-2x\ln x+2x- \frac{1}{7} x^7+ \frac{2}{5} x^5 \right]^{1.507397}_{0.447141}}{1.419562} \\ \\ = \frac{2.41459-1.64949}{1.419562} \\ \\ = \frac{0.7651}{1.419562} \approx0.539[/tex]
Therefore, the coordinates of the centroid of the area bounded by the given two curves is (0.985, 0.539)
